\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) \(4x\left(a-b\right)+6xy\left(b-a\right)\)

\(=4x\left(a-b\right)-6xy\left(a-b\right)\)

\(=\left(4x-6xy\right)\left(a-b\right)\)

\(=2x\left(2-3y\right)\left(a-b\right)\)

b) \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=3\left(2x+1\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(3-2x+5\right)\left(2x+1\right)\)

\(=\left(8-2x\right)\left(2x+1\right)\)

\(=2\left(4-x\right)\left(2x+1\right)\)

g: \(\left(3x-1\right)^2-\left(x+3\right)^2\)

\(=\left(3x-1-x-3\right)\left(3x-1+x+3\right)\)

\(=\left(2x-4\right)\left(4x+2\right)\)

\(=4\left(x-2\right)\left(2x+1\right)\)

\(b,x^3-3x^2-4x+12\)

\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-4\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

\(c,3x^3-7x^2+17x-5\)

\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)

\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)

\(\text{d) 2x}^4- 7x^3 - 2x^2 + 13x + 6\)
\(\text{= (2x^4 + 2x^3) - (9x^3 + 9x^2) + (7x^2 + 7x) + (6x + 6)}\)
\(\text{= 2x^3(x + 1) - 9x^2(x + 1) + 7x(x + 1) + 6(x + 1)}\)
\(\text{= (x + 1)(2x^3 - 9x^2 + 7x + 6)}\)
\(\text{= (x + 1)(2x + 1)(x - 3)(x - 2)}\)

phần b,c thay ''<=>'' là ''='' nhé ! Mình nhầm!

bài 1:

a) x(x-2)-5y-(x-2)=(x-5y)(x-2)

b) =(2x-3-4x)(2x-3+4x)=(-2x-3)(6x-3)

bài 2 bạn tự luyện nhé

x³-3x²-4x+12=(x³-3x²)-(4x-12)=x²(x-3)-4(x-3)=(x-3)(x²-4)=(x-3)(x-2)(x+2)

\(x^3-3x^2-4x+12=\left(x^3-3x^2\right)-\left(4x-12\right)\\ =x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)\\ =\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

\(x^3-3x^2-4x+12\)

\(=\left(x^3-3x^2\right)-\left(4x-12\right)\)

\(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-4\right)\)

\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

\(x^4-4x^3-2x^2-3x+2\)

\(\Leftrightarrow x^4+x^3-5x^3+x^2-5x^2+2x^2-5x+2x+2\)

\(\Leftrightarrow x^4+x^3+x^2-5x^3-5x^2-5x+2x^2+2x+2\)

\(\Leftrightarrow x^2\left(x^2+x+1\right)-5x\left(x^2+x+1\right)+2\left(x^2+x+1\right)\)

\(\Leftrightarrow\left(x^2-5x+2\right)\left(x^2+x+1\right)\)

Xin tick ạ !!!

#)Giải :

\(x^3-2x-4\)

\(=x^3+2x^2-2x^2+2x-4x-4\)

\(=x^3+2x^2+2x-2x^2-4x-4\)

\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)

\(=\left(x-2\right)\left(x^2+2x+2\right)\)

\(x^4+2x^3+5x^2+4x-12\)

\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)

\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)

\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)

\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)

Câu 1.

Đoán được nghiệm là 2.Ta giải như sau:

\(x^3-2x-4\)

\(=x^3-2x^2+2x^2-4x+2x-4\)

\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+2x+2\right)\)

cảm ơn nha!